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Assuming infinite speed, then the lowest number of rain hits is 12,000,000.

We want to get twice as wet, so we want the head drop counts to be 12,000,000. The rain is falling at the rate of 10,000 drops/sec/ft2. That means we need to get across the 1000 ft field in 12,000,000/10,000 = 1,200 seconds.

That works out to be 5/6 ft/sec, which is probably should be called a strolling pace rather than a running pace =)

Since you are (essentially) a box that's always in a standing position and moving through rain (assumed to be) falling straight down, you will get wet on top and on the side pointing the direction you are moving (assumed to be the largest side... no running sideways).

Like Bushindo started with, if you run infinitely fast, only the side of the box will get wet... your head will remain dry. If you imagine this face of the box as carving a space out through time, you'll see the volume will be 6 feet * 2 feet * 1000 feet = 12000 ft3 regardless of how long it takes to get there. This is because if you skew a rectangular prism parallel to two sides, the volume doesn't change.

To get twice as wet, we need the face of the box representing your head to carve out an equal area in the falling rain. I'll simplify this by ignoring the last 6 inches of the journey (your thickness), and the result is simply another skewed rectanglular prism (but one for which the distance separating the two parallel lines you are skewing along is dependant on time). So the volume/time will be 2 feet * 6 inches * 10 ft/sec = 10 ft3/sec. Since we want this volume to equal 12000 ft3, we divide them. This is 12000/10 = 1200 seconds. The needed speed is 1000 ft / 1200 seconds = 5/6 feet per second.

So it looks like Bushindo was right.

Spoiler for Yay. Let's make this unnecessarily complicated!

If we want to include the time it takes to get into the shelter (your thickness further will be needed, and the rain falling on your head will carve out a rectanglular prism cut diagonally in half). This will result in a total volume of 1000 ft / speed * 10 ft3/sec + .5 ft / speed * .5 * 10ft3/sec.

What speed would I need to run sideways to get just as wet as running infinitely fast running normally?

We already found the area carved out running infinitely fast is 12000 ft3/sec.

The area carved out by the side of the box would simply be a quarter of the 12000 ft3/sec, so 3000ft3/sec. This leaves 9000 ft3/sec for the rain falling on your head.

So using the volume/time found previously, 10 ft3/sec, we just need to divide the volume by that.9000/10 = 900 seconds. This leaves the speed at 1000/900 ft/sec = 1.111111 ft/sec. So going a little bit faster (.277777777 ft/sec, which means 33% faster) and moving sideways towards your destination will have you get half as wet.

Conclusion: Always run sideways in the rain. Hmm... or maybe crawl on your side!

Think of this in terms of extremes.
If you go super fast, you will only get 1 block of rain hitting you.
If you stand still, you will get every block of rain hitting you (in that location).

Theoretically the faster you move, the less rain should hit you.

Now, you should get the same water contact on you whether standing still for one second, or running for one second. Standing still will be more water on top of you (head), while running will be making contact with your chest.

No matter how fast or slow you move, the front of your rectangle will ALWAYS hit a rectangle of water with the size of 6x2x1000.
It is impossible for you to not hit every water droplet in front of you (assuming equal and even distribution of water droplets).

Therefore the only discrepancy is how much water hits the top of your rectangle. The slower you go, the more hits your top.
So faster is better.

Good analysis all, but it's actually faster than the answers given so far.

Spoiler for Answer

An object presenting a frontal area of F and a top area of Ttraveling a distance x at a speed sin a rain of density d and speed sr will intercept a number of raindrops D given by

D = dx [F + Tsr/s] which is minimal when s tends to infinity:

Dmin = d x F = 1,000,000 F [for the values in the OP]

D = 2 Dmin when F = T sr/s, or when s = sr T/F

The OP tells us the man does not lean, so from the given dimensions we know that T = 1.The OP does not dictate whether the man faces forward, sideways or somewhere in between.His forward orientation is not in the list of things that can be assumed.

Good analysis all, but it's actually faster than the answers given so far.

Spoiler for Answer

An object presenting a frontal area of F and a top area of Ttraveling a distance x at a speed sin a rain of density d and speed sr will intercept a number of raindrops D given by

D = dx [F + Tsr/s] which is minimal when s tends to infinity:

Dmin = d x F = 1,000,000 F [for the values in the OP]

D = 2 Dmin when F = T sr/s, or when s = sr T/F

The OP tells us the man does not lean, so from the given dimensions we know that T = 1.The OP does not dictate whether the man faces forward, sideways or somewhere in between.His forward orientation is not in the list of things that can be assumed.